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ემზადებით გამოცდისთვის? მოემზადეთ ამ 3 გაკვეთილის დახმარებით შემდეგ თემაზე: Buffers, titrations, and solubility equilibria
იხილეთ 3 გაკვეთილი
ვიდეოს აღწერა
- [Voiceover] We've been looking at the titration curve for the titration of a weak acid, acetic acid, with a strong base, sodium hydroxide. And in Part A, we found the pH before we'd added any base at all. So, we found this point on our titration curve. And we also found in Part B, the pH after you add 100 mL of base. So, we found the pH at this point. In Part C, our goal is to find the pH after the addition 200 mL of 0.500 molar solution of sodium hydroxide. So, how many moles of hydroxide ions are we adding here? So, if the concentration of sodium hydroxide is 0.0500 M, that's the same concentration for hydroxide. So, for hydroxide, we have the concentration of 0.0500 M, and molarity is moles over liters, right? So moles over liters. So, we have 200 mL, and if we move our decimal place one, two, three, that's 0.2 liters. All right, so this is 0.2000 liters here. So, we solve for moles, so 0.05 times 0.2 is going to give us 0.0100, 0.0100 moles of hydroxide ions. So, that's how many moles of hydroxide ions we're adding to our original acid solution. How many moles of acid did we start with? Well, we had 50 mL of a 0.200 molar solution of acetic acid. So, the concentration of acetic acid, the concentration of acetic acid was equal to 0.200 M, and that's equal to moles over liters. So, 50.0 mL would be 0.0500 liters. So this would be 0.0500 liters. Once again, solve for moles. So, 0.200 times 0.0500 gives us 0.0100 moles of acetic acids. So notice we have the same number of moles of acid as we do of base. And the base is going to neutralize the acid. Let's go ahead and write the reaction. Let's write the neutralization reaction. If we have, we start with some acetic acid. And to our acidic solution we add our sodium hydroxide. So, we're adding some sodium hydroxide here. The hydroxide ions are going to take the acidic proton. So, a hydroxide ion takes this acidic proton right here. H plus and OH minus give us H2O. If you take away the acidic proton from acetic acid, you're left with acetate, CH 3 COO minus. And we're starting with .01 moles of acetic acid. So, let's color coordinate here. So, we're starting with 0.0100 moles. Over here we'll put moles. So, 0.0100 moles of acetic acid. And that's the same number of moles of base. 0.0100, so we have 0.0100 moles of base. Notice our mol ratio is one to one, so we have one to one here. So, all of the base is going to react. It's going to completely neutralize the acid that we originally had present. So, all of our base reacts, and we end up with zero here, and all of our acid has been completely neutralized. We lose all of this. So, we lose all of that, and so we've neutralized all of our acid too. So, this is our equivalence point for this titration. And if we're losing acetic acid, we're converting acetic acid into acetate. So, if we think about starting with zero moles of acetate, and we lose 0.0100 moles of acetic acid, that turns into acetate. So, we have to write plus 0.0100 over here, so we're making that. So, we end up with 0.0100 moles of our acetate anion. All right, next. If we have moles of our acetate, and we have a concentration. So, if we find the total volume of our solution, we can find the concentration of acetate anions. So, let's do that next. We'll go back up here. What is the total volume now? We started with 50 mL, and we added 200 more. So, 50.0 and 200.0 give us 250.0 mL. So, that's our total volume now. And 250.0 mL would be, move our decimal place three, 0.25 liters. So, next we find our concentration of acetate. So, what is our concentration of acetate now? It would be moles over liters, so 0.0100. So we have 0.0100 moles over 0.2500 liters, over 0.2500 liters. So, we can go ahead and find the concentration. You can do this in your head or I'll just show you on the calculator, .01 divided by .25. It's going to give us a concentration of 0.04. So, the concentration is 0.0400 M. So, we have acetate anions in solution at the equivalence point. And acetate reacts with water. So, let's go ahead and show that reaction. So, we have acetate, which can react with water. And this reaction will eventually come to an equilibrium here. So the acetate anion acts as a base and takes a proton from water. So, if acetate picks up a proton, it turns into acetic acid. So, CH3COOH. If we take a proton away from water, we're left with hydroxide, so OH minus. So, now let's think about our initial concentration of acetate. So, our initial concentration of acetate was 0.0400. So, we write here initial concentration is 0.0400 M. And we're assuming that we don't have any products yet, so we write zero here. Next we think about our change. Well, a certain concentration of acetate is going to react, so we're going to lose a certain concentration, which we call X. Whatever we lose for acetate, we gain for acetic acid. So if it's minus X for acetate, it must be plus X over here for acetic acid. And therefore, also, plus X for hydroxide. All right, so at equilibrium... At equilibrium, our concentration would be 0.0400 minus X for acetate. And then we'd have X over here and X over here. So, if we write an equilibrium expression for this, acetate is acting as a base. So, for our equilibrium expression, we would write K b. And remember, that's concentration of products over reactants. So that would be X times X, so X times X, all over 0.0400 minus X, leaving water out. So, this is over 0.0400 minus X. And we did all this in weak base equilibrium, so make sure that you watched that video before you watch this one, because I have to go a little bit faster here. All right, next we need to find the Kb value. We can get the Kb by... Because we know the Ka. In the last video, the Ka for acetic acid was 1.8 times 10 to the negative 5. And we know Ka times Kb for a conjugate acid-based pair is equal to 1.0 times 10 to the negative 14. Again, this is from an earlier video. So, if we plug in Ka into here, we can solve for Kb. And I won't do it, to save time, on the calculator. I'll just give you the answer, that Kb is equal to 5.6 times 10 to the negative 10. So, we plug this in to our equilibrium expression, so we get 5.6 times 10 to the negative 10 is equal to X squared over... Here we make the assumption, like we did in all the earlier videos, that X is a very small number, it's a very small concentration. So, 0.0400 minus X is approximately the same thing as 0.0400. So, we write 0.0400 in here. And next we need to solve for X. So, let's take out the calculator. So we have 5.6 times 10 to the negative 10. We need to multiply that by .04. And then we need to take the square root of our answer. And so we get X is equal to 4.7 times 10 to the negative 6. So, X is equal to 4.7 times 10 to the negative 6. What does X represent? We go up here, and we notice that X represents the concentration of hydroxide ions. So this is equal to... This is equal to the concentration of hydroxide ions. So 4.7 times 10 to the negative 6 molar. All right, our goal was to find the pH. So, at this point it makes sense to find the pOH. pOH is equal to the negative log of the concentration of hydroxide ions. So we plug that into here, and we solve for the pOH. Negative log of 4.7 times 10 to the negative 6 give us a pOH of 5.33. So, the pOH is equal to 5.33. And one more step. pH plus pOH is equal to 14.00. So, if we plug in our pOH into here, pH is equal to 14.00 minus 5.33, which is 8.67. So, we're at the equivalence point, but this is a titration of a weak acid with a strong base. And so, we have a basic salt solution at the equivalence point. So, our pH is in the basic range. It's above seven. So, let's find that point on our titration curve. So, we've added here 200 mL of our base. And the pH was 8.67. So we've added 200.00 mL of our base, and the equivalence point should be somewhere in there. So right about there, about 8.67. So, that's our equivalence point for a titration of a weak acid with a strong base for this particular example. Finally we're on to Part D, which asks us, what is the pH after the addition of 300.0 mL of a 0.0500 molar solution of sodium hydroxide? So, once again, we need to find the moles of hydroxide ions that we are adding. The concentration would be equal to 0.0500, so 0.0500 molar is our concentration of hydroxide ions. And that's equal to moles over liters. So, 300.0 mL would be 0.3000 liters. So, we have 0.3000 liters here. Multiply 0.0500 by 0.3000, and you get moles. So, 0.0500 times 0.300 is equal to 0.0150 moles of hydroxide ions. In Part C, we saw that we needed 0.0100 moles of hydroxide ions to completely neutralize the acid that we originally had present. So, we're going to use up... We're going to use up 0.0100 moles of hydroxide. That's how much was necessary to neutralize our acid. All right, so how many moles of hydroxide are left over after the neutralization? Well, that would just be 0.0050. So, 0.0050 mol of hydroxide ions are left over after all the acid has been neutralized. So, our goal is to find the pH. So, we could find the pOH if we found if we found the concentration of hydroxide. So, what is the concentration of hydroxide ions now after the neutralization has occurred? So, concentration is moles over liter, so it's 0.0050 divided by... What's our total volume? We started with 50.0, and we have now added 300.0 mL more, so 300.0 plus 50.0 is 350.0 mL. Or 0.3500 liters. So, what is our concentration of hydroxide ions? So, this is .005 divided by .35. So our concentration of hydroxide ions is 0.014. So, let's write 0.014 M here. Once we know that, we can calculate the pOH. So the pOH is the negative log of the concentration of hydroxide ions. So, it's the negative log of 0.014. So, we can do that on our calculator. Negative log of .014. And we get a pOH of 1.85. All right, so our pOH. Let's get a little bit more room here. pOH is equal to 1.85. And finally, to get the pH, we know that pH plus pOH is once again equal to 14.00. So, we plug in the pOH into here. And the pH would be equal to 14.00 minus 1.85. So 14.00 minus 1.85 gives us 12.15. So, we're now past the equivalence point. All right, so we're past the equivalence point here. So our pH is 12.15. After we've added 300.0 mL of our base. So, let's find the point on our titration curve. We've added 300.0 mL of our base, so we're right here. So, we go up to our titration curve, and that would be right here. So, we just found the pH is a little bit over 12. So approximately, we've got 12.15. So that's the pH right here on our titration curve.